A-b)3+(b-c)3+(c-a)3:(a-b)(b-c)(c-a)

The answer to the above equation is 3

Step-by-step explanation:

(a-b)³+(b-c)³+(c-a)³: (a-b)(b-c)(c-a)

Let us consider  (a−b)= x, (b−c)= y and (c−a)= z.  

Hence, It is obvious that:

x+y+z =0 ∵all the terms gets cancelled out

⇒We must remember the algebraic formula

x³+y³+z³−3xyz= (x+y+z) (x²+y²+z²-xy-xz-yz)

Since x+y+z=0 ⇒Whole “(x+y+z) (x²+y²+z²-xy-xz-yz) ” term becomes 0

x³+y³+z³−3xyz =0

Alternatively, x³+y³+z³= 3xyz

Now putting the value of x, y, z in the original equation

(a-b)³+(b-c)³+(c-a)³ can be written as 3(a-b)(b-c)(c-a) since  (a−b)= x, (b−c)= y and (c−a)= z.

3(a-b)(b-c)(c-a): (a-b)(b-c)(c-a)

= 3           ∵Common factor (a-b)(b-c)(c-a) gets cancelled out

Answer to the above question is 3

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.

2 2/3 hours duhh silly

step-by-step explanation: